（持续更新ing...）
式子没啥可说的，直接列式子吧（证明都在最下面）：
(资料图片)
\(1. \displaystyle \sum_{i = 1}^n i^2 = \frac{n(n + 1)(2n + 1)}{6}\)
\(2. \displaystyle \sum_{1 \le i < j \le n} (i + j) = \frac{n(n - 1)(n + 1)}{2}\)
\(3. \displaystyle \sum_{1 \le i \le j \le n} (i + j) = \frac{n^2(n + 1)}{2}\)
\(4. 对于某一等比数列，有 \displaystyle S_n = a_1 \cdot \frac{1 - q^n}{1 - q} (S_n为前n项的和，q 为公比)\)
证明：$1. $
\[\begin{aligned}\sum_{i = 1}^n i^2&= \sum_{i = 1}^n \sum_{j = 1}^i i\\\sum_{i = 1}^n i^2&= \sum_{1 \le j \le i \le n}^n i\\\sum_{i = 1}^n i^2&= \sum_{j = 1}^n \sum_{i = j}^n i\\\sum_{i = 1}^n i^2&= \sum_{j = 1}^n \frac{(n + j)(n - j + 1)}{2}\\\sum_{i = 1}^n i^2&= \frac{1}{2} \cdot \sum_{j = 1}^n (n^2 - j^2 + n + j)\\\sum_{i = 1}^n i^2&= \frac{1}{2} \cdot \sum_{j = 1}^n (n(n + 1) - j^2 + j))\\2 \cdot \sum_{i = 1}^n i^2&= \sum_{j = 1}^n (n(n + 1)) - \sum_{j = 1}^n j^2 + \sum_{j = 1}^n j\\3 \cdot \sum_{i = 1}^n i^2&= \sum_{j = 1}^n (n(n + 1)) + \sum_{j = 1}^n j\\3 \cdot \sum_{i = 1}^n i^2&= n^2(n + 1) + \frac{n(n + 1)}{2}\\3 \cdot \sum_{i = 1}^n i^2&= \frac{2n \cdot n(n + 1) + n(n + 1)}{2}\\3 \cdot \sum_{i = 1}^n i^2&= \frac{n(n + 1)(2n + 1)}{2}\\\sum_{i = 1}^n i^2&= \frac{n(n + 1)(2n + 1)}{6}\\\end{aligned}\]$2. $
\[\begin{aligned}\sum_{1 \le i < j \le n} (i + j)&= \sum_{i = 1}^n \sum_{j = 1}^{i - 1} (i + j)\\&= \sum_{i = 1}^n (i(i - 1) + \frac{i(i - 1)}{2})\\&= \frac{3}{2} \cdot \sum_{i = 1}^n i(i - 1)\\&= \frac{3}{2} \cdot (\sum_{i = 1}^n i^2 - \sum_{i = 1}^n i)\\&= \frac{3}{2} \cdot (\frac{n(n + 1)(2n + 1)}{6} - \frac{n(n + 1)}{2})\\&= \frac{3}{2} \cdot \frac{n(n + 1)(2n - 2)}{6}\\&= \frac{n(n - 1)(n + 1)}{2}\\\end{aligned}\]有一个地方用到了第一个式子的结论。
$3. $ 同第二个式子的证明，只不过加上了 \(\displaystyle \frac{n(n + 1)}{2}\)
$4. $
\[\begin{aligned}S_n&= a_1 + a_2 + \dots + a_{n - 1} + a_n \\q \cdot S_n&= q \cdot a_1 + q \cdot a_2 + \dots + q \cdot a_{n - 1} + q \cdot a_n = S_{n + 1} \\S_n - q \cdot S_n &= (1 - q) \cdot S_n = S_n - S_{n + 1} = a_1 - a_{n + 1} \\a_{n + 1} &= a_1 \cdot q^n \\S_n &= a_1 \cdot \frac{1 - q^n}{1 - q}\end{aligned}\]标签：
